r"""
给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。
说明: 叶子节点是指没有子节点的节点。

示例: 
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。

作者：力扣 (LeetCode)
链接：https://leetcode-cn.com/leetbook/read/data-structure-binary-tree/xo566j/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
"""


class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if not root:
            return False
        if root.val == sum and root.left is None and root.right is None:
            return True
        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)


if __name__ == '__main__':
    tree = TreeNode(5)
    tree.left = TreeNode(4)
    tree.right = TreeNode(8)
    tree.left.left = TreeNode(11)
    tree.left.left.left = TreeNode(7)
    tree.left.left.right = TreeNode(2)
    tree.right.left = TreeNode(13)
    tree.right.right = TreeNode(4)
    tree.right.right.right = TreeNode(1)
    tree1 = TreeNode(-2)
    tree1.right = TreeNode(-3)
    s = Solution()
    print(s.hasPathSum(tree, 22))
    print(s.hasPathSum(tree1, -5))
